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Let $R$ be a commutative ring with unity such that $n!$ is not a zero-divisor. Let $s_1=\sigma_1,s_2,s_3\cdots$ and $\sigma_1,\sigma_2\cdots$, (convention: if $k>n$, then $\sigma_k=0$) be elements of $R$. We consider the Newton's identities

$s_k=\sigma_1s_{k-1}-\cdots+(-1)^k\sigma_{k-1}s_1+(-1)^kk\sigma_k$.

We assume that there is a zero sequence of length $n$: $s_p=\cdots=s_{p+n-1}=0$. Formal calculations seem to indicate that the $(\sigma_i)_{1\leq i\leq n}$ are in the nilradical of $R$. In particular, the result is true for $(n=6,p\leq9),(n=7,p\leq8),(n=8,p\leq 7),(n=9,p\leq5)$.

Is this result true in general ?

Thanks in advance.

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  • $\begingroup$ You might be interested in a variation presented here: mathoverflow.net/questions/155114/… . It represents something close to your form as a product, and you might find in it inspiration for a proof that terms for large enough k are in the nilradical. Gerhard "Ask Me About Myopic Mathematics" Paseman, 2014.04.29 $\endgroup$
    – Gerhard Paseman
    Commented Apr 29, 2014 at 18:48

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If that helps, in Section III.2 of my article http://link.springer.com/article/10.1007/s00023-003-0127-7 there is a proof due to de Calan and Magnen of the fact: for any $k\ge 1$, $({\rm tr}\ N)^{k(n-1)+1}$ is in the ideal generated by the matrix elements of $N^k$. Here $N$ is an $n\times n$ matrix of indeterminates.

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    $\begingroup$ See also two (I believe) proofs in Doron Zeilberger, Gert Almkvist's generalization of a mistake of Bourbaki, math.rutgers.edu/~zeilberg/mamarim/mamarimhtml/gert.html . $\endgroup$
    – darij grinberg
    Commented Apr 29, 2014 at 20:12
  • $\begingroup$ Thanks Abdel. Do you know a proof of this fact ? $\endgroup$
    – loup blanc
    Commented May 1, 2014 at 21:17
  • $\begingroup$ Darij, that is a very interesting reference. If you want the green chevron, then post your comment as an answer. $\endgroup$
    – loup blanc
    Commented May 1, 2014 at 21:19
  • $\begingroup$ @loup blanc: were you paying attention? the paper I mentioned contains a proof of that fact. $\endgroup$
    – Abdelmalek Abdesselam
    Commented May 2, 2014 at 13:29
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    $\begingroup$ I see. The arxiv version is quite different. The proof is in the published version indicated in the link above. If you do not have access to the journal and email me I can send you a copy. $\endgroup$
    – Abdelmalek Abdesselam
    Commented May 2, 2014 at 20:46

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